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In triangle ABC, points D and E lie on BC and AC, respectively.

If AD and BE intersect at T so that AT/DT = 3 and BT/ET = 4, what is CD/BD?
 

$\text{Let $\vec{CA}=\vec{b}$} \\ \text{Let $\vec{CB}=\vec{a}$} \\ \text{Let $\mu = \dfrac{3}{4} = \dfrac{AT}{AD}$} \\ \text{Let $1-\mu = \dfrac{1}{4} = \dfrac{DT}{AD}$} \\ \text{Let $\rho = \dfrac{4}{5} = \dfrac{BT}{BE}$} \\ \text{Let $1-\rho = \dfrac{1}{5} = \dfrac{ET}{BE}$} \\ \text{Let $\lambda = \dfrac{BD}{BC} $} \\ \text{Let $1-\lambda = \dfrac{CD}{BC} $} \\ \text{Let $\mathbf{ \dfrac{1-\lambda}{\lambda} = \dfrac{CD}{BD}} $} \\ \text{Let $\epsilon = \dfrac{CE}{CA} $}$

$\text{In C-D-T-E} \\ \begin{array}{|rcll|} \hline (1-\lambda)\vec{a}+(1-\mu)[\vec{b}-(1-\lambda)\vec{a}] &=& \epsilon \vec{b} + (1-\rho)(\vec{a}-\epsilon \vec{b}) \\ (1-\lambda)\vec{a}+(1-\mu)\vec{b} -(1-\mu)(1-\lambda)\vec{a} &=& \epsilon \vec{b} + (1-\rho)\vec{a} - (1-\rho)\epsilon \vec{b}) \\ \vec{a}[(1-\lambda)-(1-\mu)(1-\lambda)-(1-\rho) ] &=& \vec{b} [\epsilon - (1-\rho)\epsilon -(1-\mu) ] \\ \vec{a}[(1-\lambda)\Big(1-(1-\mu)\Big)-(1-\rho) ] &=& \vec{b} [\epsilon - (1-\rho)\epsilon -(1-\mu) ] \\ \vec{a}[\underbrace{ (1-\lambda)\mu)-(1-\rho) }_{=0} ] &=& \vec{b} [\underbrace{ \epsilon - (1-\rho)\epsilon -(1-\mu) }_{=0} ] \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline (1-\lambda)\mu-(1-\rho) &=& 0 \\ (1-\lambda)\mu &=& (1-\rho) \\\\ \mathbf{ 1-\lambda } &=& \mathbf{\dfrac{(1-\rho)}{\mu} } \\ \hline 1-\lambda &=& \dfrac{(1-\rho)}{\mu} \\ \lambda &=& 1-\dfrac{(1-\rho)}{\mu} \\ \mathbf{ \lambda} &=& \mathbf{ \dfrac{\mu-(1-\rho)}{\mu} } \\ \hline \dfrac{CD}{BD} &=& \dfrac{1-\lambda}{\lambda} \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{(1-\rho)}{\mu}} {\dfrac{\mu-(1-\rho)}{\mu}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{(1-\rho)} {\mu-(1-\rho)} \\ \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{1}{5}} {\dfrac{3}{4}-\dfrac{1}{5}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{1}{5}} {\dfrac{11}{20}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{1}{5} * \dfrac{20}{11} \\ \\ \mathbf{ \dfrac{CD}{BD}} &=& \mathbf{\dfrac{4}{11}} \\ \hline \end{array}$

Find the number of paths from P to Q, where each step is to the right or up.

In general:

$\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{n} \\\\ \mathbf{x+y } &\ge& \mathbf{4n} \\ \hline \end{array}$

$\text{The smallest possible value for $x+y$ is $4n$}$

Given positive integers $x$ and $y$ such that $x \ne y$ and $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{18}$,
what is the smallest possible value for $x+y$?

$\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{18} \\\\ \dfrac{x+y}{xy} &=& \dfrac{1}{18} \\\\ \mathbf{xy} &=& \mathbf{18*(x+y)} \\ \hline \end{array}$

$\large{AM\ge GM}$

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{x+y}{2}} &\ge& \mathbf{\sqrt{xy}} \\ x+y &\ge& 2\sqrt{xy} \quad &| \quad \text{square both sides} \\ (x+y)^2 &\ge& 4xy \quad | \quad xy = 18*(x+y) \\ (x+y)^2 &\ge& 4*18*(x+y) \\ x+y &\ge& 4*18 \\ \mathbf{x+y } &\ge& \mathbf{72} \\ \hline \end{array}$

$\text{The smallest possible value for $x+y$ is $72$}$

Find all positive integers n, less than 17, for which n! is an integral multiple of 49.

$\boxed{\dfrac{n}{7}=2 \\ \mathbf{n}=\mathbf{14} }$

For a certain value of $k$, the system
$\begin{align*} x + y + 3z &= 10, \\ -4x + 3y + 5z &= 7, \\ kx + z &= 3 \end{align*}$
has no solutions. What is this value of $k$?

$\begin{array}{|rcll|} \hline \begin{vmatrix} 1 & 1 & 3 \\ -4 & 3 & 5 \\ k & 0 & 1 \\ \end{vmatrix} &=& 0 \\\\ k* \begin{vmatrix} 1 & 3 \\ 3 & 5 \\ \end{vmatrix} +1* \begin{vmatrix} 1 & 1 \\ -4 & 3 \\ \end{vmatrix} &=&0 \\\\ k*(1*5-3*3) + 1*(1*3-(-4)*1) &=& 0 \\\\ k*(-4) + 1*7 &=& 0 \\\\ k*(-4) &=& -7 \\\\ k &=& \dfrac{-7}{-4} \\\\ \mathbf{k} &=& \mathbf{ \dfrac{7}{4} } \\ \hline \end{array}$

When n is divided by 10, the remainder is a.
When n is divided by 11, the remainder is b.
What is n modulo 110, in terms of a and b?

$\begin{array}{|rcll|} \hline n &\equiv& {\color{red}a} \pmod{10} \quad \text{or} \quad n = a+10r,~ r\in\mathbb{Z} \\ n &\equiv& {\color{red}b} \pmod{11} \quad \text{or} \quad n = b+11s,~ s\in\mathbb{Z} \\ n &\equiv& {\color{red}x} \pmod{110} \quad |\quad 110 = 10*11\\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline &n &=& a+10r \quad | \quad *11 \\ &\mathbf{11n} &=& \mathbf{11a+10*11r} \qquad (1) \\\\ &n &=& b+11s \quad | \quad *10 \\ &\mathbf{10n} &=& \mathbf{10b+10*11s} \qquad (2) \\ \hline (1)-(2): & 11n - 10n &=& 11a+10*11r - 10b - 10*11s \\ & n &=& 11a - 10b +10*11(r-s) \quad | \quad \text{let } r-s = t\\ & n &=& 11a - 10b +10*11t \quad |\quad 110 = 10*11\\ & \mathbf{n} &=& \mathbf{11a - 10b +110t} \qquad (3) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{n} &=& \mathbf{11a - 10b +110t} \qquad (3) \\ n &=& \underbrace{\color{red}11a-10b}_{=x} \pmod{110} \\ \hline \end{array}$

$n \pmod{110} \equiv 11a-10b$

Let a and b with a > b > 0 be real numbers satisfying $a^3 + b^3 = a + b$.
Find $\dfrac{(a + b)^2 - 1}{ab}$.

$Formula: \boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}$

$\begin{array}{|rcll|} \hline a^3+b^3 &=& (a+b)(a^2-ab+b^2) \quad | \quad a^3 + b^3 = a + b \\ (a+b) &=& (a+b)(a^2-ab+b^2) \\ a^2-ab+b^2 &=& \dfrac{a+b}{a+b} \\ a^2-ab+b^2 &=& 1 \\ a^2+b^2-ab &=& 1 \quad | \quad a^2+b^2=(a+b)^2-2ab \\ (a+b)^2-2ab-ab &=& 1 \\ (a+b)^2-3ab &=& 1 \\ (a+b)^2-1 &=& 3ab \quad | \quad :ab \\\\ \mathbf{ \dfrac{ (a+b)^2-1 }{ab} } &=& \mathbf{ 3 } \\ \hline \end{array}$

The smallest possible value of n is a four-digit positive integer

$\begin{array}{|rcll|} \hline \mathbf{n} &=& \mathbf{59 + 315k} \\ 59 + 315k &>& 999 \\ 315k &>& 999 - 59 \\ 315k &>& 940 \\\\ k &>& \dfrac{940}{315} \\\\ k &>& 2.9841269841 \\ \mathbf{k} &=& \mathbf{3} \\ n_{\text{The smallest possible value of n is a four-digit positive integer }} &=& 59 + 315*3 \\ \mathbf{n} &=& \mathbf{1004} \\ \hline \end{array}$

n is a four-digit positive integer.
Dividing n by 9, the remainder is 5.
Dividing n by 7, the remainder is 3.
Dividing n by 5, the remainder is 4.
What is the smallest possible value of n?

$\begin{array}{|rcll|} \hline n &\equiv& {\color{red}5} \pmod{9} \\ n &\equiv& {\color{red}3} \pmod{7} \\ n &\equiv& {\color{red}4} \pmod{5} \\ \text{Let}~ m &=& 9*7*5 = 315 \\ \hline \end{array}$

Because 9 and 7 and 5 are relatively prim $\Big(\gcd(9,7,5)=1\Big)$,
we can go on.

$\begin{array}{|rcll|} \hline n &=& {\color{red}5} *7*5*\dfrac{1}{7*5}\pmod{9} \\ && +{\color{red}3} *9*5* \dfrac{1}{9*5}\pmod{7} \\ && +{\color{red}4} *9*7* \dfrac{1}{9*7}\pmod{5} \\ && + 315k \quad | \quad k \in \mathbb{Z} \\\\ n &=& 175* \left(\dfrac{1}{35}\pmod{9}\right) \\ && +135*\left(\dfrac{1}{45}\pmod{7}\right) \\ && +252* \left(\dfrac{1}{63}\pmod{5}\right) \\ && + 315k \\\\ \hline \end{array} \begin{array}{|lcll|} \hline \dfrac{1}{35}\pmod{9} \quad | \quad 35 \equiv -1 \pmod{9} \\ \equiv \dfrac{1}{-1}\pmod{9} \\ \dfrac{1}{35}\pmod{9}\equiv -1\pmod{9} \\ \hline \dfrac{1}{45}\pmod{7} \quad | \quad 45 \equiv 3 \pmod{7} \\ \equiv \dfrac{1}{3}\pmod{7} \\ \equiv 3^{\phi(7)-1}\pmod{7} \\ \equiv 3^{6-1}\pmod{7} \\ \equiv 3^{5}\pmod{7} \\ \equiv 243 \pmod{7} \\ \dfrac{1}{45}\pmod{7} \equiv 5 \pmod{7} \\ \hline \dfrac{1}{63}\pmod{5} \quad | \quad 63 \equiv 3 \pmod{5} \\ \equiv \dfrac{1}{3}\pmod{5} \\ \equiv 3^{\phi(5)-1}\pmod{5} \\ \equiv 3^{4-1}\pmod{5} \\ \equiv 3^{3}\pmod{5} \\ \equiv 27 \pmod{5} \\ \dfrac{1}{63}\pmod{5}\equiv 2 \pmod{5} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline n &=& 175* \left(\dfrac{1}{35}\pmod{9}\right) \\ && +135*\left(\dfrac{1}{45}\pmod{7}\right) \\ && +252* \left(\dfrac{1}{63}\pmod{5}\right) \\ && + 315k \\\\ n &=& 175*(-1) +135*5 +252*2 + 315k \\ n&=& -175 +675 + 504 +315k \\ n &=& 1004 + 315k \quad | \quad 1004 \equiv 59 \pmod{315} \\ n &=& 59 + 315k \\ \mathbf{ n_{\text{min.}}} &=& \mathbf{ 59 } \\ \hline \end{array}$